[Anne Wanyonyi]

Thank you Sebastian.

I am now in panic mode.
I have 40 observations. I had already carried out the bounds test procedure and it points to existence of cointegration. So that means, there is a high chance the conclusion I made from the bounds test is incorrect?
All my variables are in log form.

[Sebastian Kripfganz]

Zuhura Anne: You could try Kit Baum's cusum6 package:

ssc describe cusum6

I have never used it and thus cannot give any further advice on it. Please consult its help file to find out how to use it.

[Sebastian Kripfganz]

Anne Wanyonyi: Don't panic! If there is only weak evidence for non-normality of the errors, you could still use the bounds test and maybe add a comment in your work that the results need to be interpreted with caution. If there is strong evidence for non-normality, you might have to go back to the drawing board and think about your model specification. If you have not yet included a time trend, this could be worth trying.

[Anne Wanyonyi]

Sebastian, I have gone back and looked at my data again.
All my variables are I(0) or I(1) when I test for unit root using ADF. When I apply the unit root tests with breaks not all are I(1) or I(0). Does this invalidate the use of ARDL?

Lastly, how do I incorporate the time trend in the ARDL?

[Sebastian Kripfganz]

The bounds test clearly requires that all variables are I(0) or I(1). If you have structural breaks, you might want to directly model them. Otherwise, your model is likely to be misspecified. Depending on the nature of the breaks, you could for example include dummy variables for a specific time period with the exog() option of ardl. For the the bounds test to be still valid in the presence of such dummy variables, the time period that is captured by them should be limited. For example, you could specify a dummy variable that takes on the value 1 for a small number of periods if these periods are structurally different such that they cause a temporary break in the time series. The bounds test is not suitable in the presence of permanent structural breaks.

[Zuhura Anne]

I have tried the cusum6 and from the output, I do not have parameter stability. How can I address the issue of parameter stability?
In addition, if some variables are I(2) which model would you advise one to use?

Thank you very much Sebastian for the timely responses.

[Anne Wanyonyi]

If you have structural breaks, you might want to directly model them

Kindly what do you mean by this?
How can I distinguish between a permanent and temporary break in the series?

[Sebastian Kripfganz]

Dear both,
If the breaks are known due to some identifiable events, you can model them with dummy variables that shift the intercept for some time period. By interacting these dummies with the regressors, you can model breaks in the short-run coefficients. Similarly, you can interact the dummy variables with the trend variable to allow for different trends in different periods. It really depends on the type of break. Another possibility could be to split the time series if you have enough observations. I could already be sufficient to remove a few observations at the beginning or end (for example, use only observations prior to the financial crisis) and see if this gives you parameter stability.

A temporary break could be a shift that lasts for a couple of time points only (some crisis period) but afterwards the time series' behave in the same way as before this period.

I am not an expert on structural breaks / parameter stability and cannot provide more specific help on how to identify them or how to deal with them.

[Zuhura Anne]

Thank you so much for the responses. The trend and dummy variables have been so helpful. I have parameter stability and all other diagnostic tests are satisfied. However, normality is not upheld when I use this syntax:

mvtest normality variable

Is this something to worry about or since most of the properties have been satisfied, then I can continue using the model?

[Sebastian Kripfganz]

It is not really possible to give a general answer to this question. Please also see my comments #269 and #273 above.

[NIP NIP]

Thanks!

[Vernon Mila]

Some questions (please see attachment for full question with images of outputs included)>>

1. Here is the data we’re looking to run ARDL on – LunitedStates is the dependent variable and LUnitedStatesPS is independent variable
   1. Given below plot would we run the following specs for ARDL?
      1. ardl LUnitedStates LUnitedStatesPS, lags(3 1) ec [constant] trendvar(LUnitedStates)
   2. When I run above I get below error –
      1. Trend variable must be collinear with the time variable.
      2. I maybe mis-specifying above – how may I correctly specify this model
2. How do I tell the ardl to explicitly include contemeraneous terms in the independent variables (X’s) or is this always a default (i.e. LAG 0)?
   1. For eg. ardl LUnitedStates LUnitedStatesPS, lags(3 1), automatically includes lag 0
   2. When I am modeeling ec form, ardl LUnitedStates LUnitedStatesPS, lags(3 1) ec, only has the contemperanous difference in the “short run” equation due to inclusion of only Lag(1)
   3. See outputs below – should I worry if PART (b) is not including the contemperaneous term?

[Daniel Schneider]

The previous post depends heavily on a Word attachment which is discouraged in Section "12.5 Posting attachments: please don't..." in the Statalist FAQ. Since it would be difficult to follow the discussion, I have responded to the inquiry in a private email.

[Dimas Bagas Herlambang]

Sebastian,

I want to ask about F-test for PSS. I run ardl in ec1 mode, and got this result:

. ardl lr mrpos mrneg lnlipos lnlineg, ec1 fast lags(. . . . .) maxlags(6 6 6 6 6) maxcombs(15000)

ARDL regression
Model: ec

Sample:  2000m7 -  2017m2
Number of obs  = 200
Log likelihood = 122.64809
R-squared      = .51863377
Adj R-squared  = .50108396
Root MSE       = .13375199

------------------------------------------------------------------------------
        D.lr |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
ADJ          |
          lr |
         L1. |  -.0506548   .0122413    -4.14   0.000    -.0747994   -.0265101
-------------+----------------------------------------------------------------
LR           |
       mrpos |
         L1. |   .3373259   .0836383     4.03   0.000     .1723581    .5022938
             |
       mrneg |
         L1. |   .3748366   .0895931     4.18   0.000     .1981234    .5515498
             |
     lnlipos |
         L1. |  -1.291288   .3078728    -4.19   0.000    -1.898535   -.6840407
             |
     lnlineg |
         L1. |  -1.295418   .3102576    -4.18   0.000    -1.907368   -.6834666
-------------+----------------------------------------------------------------
SR           |
          lr |
         LD. |   .4615063   .0709036     6.51   0.000     .3216562    .6013564
        L2D. |   .2238202   .0698491     3.20   0.002     .0860501    .3615902
             |
       mrpos |
         D1. |   .0170872   .0054116     3.16   0.002     .0064134    .0277609
             |
       mrneg |
         D1. |   .0189873   .0061479     3.09   0.002     .0068613    .0311133
             |
     lnlipos |
         D1. |  -.0654099   .0237949    -2.75   0.007    -.1123428   -.0184769
             |
     lnlineg |
         D1. |  -.0656191   .0240508    -2.73   0.007    -.1130568   -.0181813
             |
       _cons |    1.37524   .4237449     3.25   0.001     .5394466    2.211033
------------------------------------------------------------------------------

after that, I try using estat btest and test to compare the PSS F-stat, and the results are:

. estat btest

Pesaran/Shin/Smith (2001) ARDL Bounds Test
H0: no levels relationship             F =  3.620
                                       t = -4.138


. test L1.lr = L1.mrpos = L1.mrneg = L1.lnlipos = L1.lnlineg = 0

 ( 1)  [ADJ]L.lr - [LR]L.mrpos = 0
 ( 2)  [ADJ]L.lr - [LR]L.mrneg = 0
 ( 3)  [ADJ]L.lr - [LR]L.lnlipos = 0
 ( 4)  [ADJ]L.lr - [LR]L.lnlineg = 0
 ( 5)  [ADJ]L.lr = 0

       F(  5,   192) =   30.74
            Prob > F =    0.0000

Does it meant to be different? and which one represent the true F-stat according to PSS?

Thank you

[Sebastian Kripfganz]

You cannot compute the F-statistic directly from the error-correction output. Please see slides 9 and 10 of my presentation at the 2016 Stata Conference in Chicago.

You can recover it from the ARDL representation (ardl without the ec1 option), e.g.

. webuse lutkepohl2
(Quarterly SA West German macro data, Bil DM, from Lutkepohl 1993 Table E.1)

. ardl ln_inv ln_inc, lags(2) ec1

ARDL regression
Model: ec

Sample: 1960q3 - 1982q4
Number of obs  = 90
Log likelihood = 158.62362
R-squared      = .13799034
Adj R-squared  = .08668024
Root MSE       = .04298403

------------------------------------------------------------------------------
    D.ln_inv |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
ADJ          |
      ln_inv |
         L1. |  -.1515065   .0624739    -2.43   0.017    -.2757427   -.0272702
-------------+----------------------------------------------------------------
LR           |
      ln_inc |
         L1. |   .8237427   .0599558    13.74   0.000      .704514    .9429713
-------------+----------------------------------------------------------------
SR           |
      ln_inv |
         LD. |  -.1362943   .1078979    -1.26   0.210     -.350861    .0782725
             |
      ln_inc |
         D1. |    .565835    .402307     1.41   0.163    -.2341965    1.365867
         LD. |    .655617   .3925048     1.67   0.099     -.124922    1.436156
             |
       _cons |   .0302675   .0666281     0.45   0.651    -.1022298    .1627648
------------------------------------------------------------------------------

. estat btest

Pesaran/Shin/Smith (2001) ARDL Bounds Test
H0: no levels relationship             F =  3.083
                                       t = -2.425
[some output omitted]

. ardl ln_inv ln_inc, lags(2)

ARDL regression
Model: level

Sample: 1960q3 - 1982q4
Number of obs  = 90
Log likelihood = 158.62362
R-squared      = .99160203
Adj R-squared  = .99110215
Root MSE       = .04298403

------------------------------------------------------------------------------
      ln_inv |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
      ln_inv |
         L1. |   .7121993   .1083168     6.58   0.000     .4967993    .9275992
         L2. |   .1362943   .1078979     1.26   0.210    -.0782725     .350861
             |
      ln_inc |
         --. |    .565835    .402307     1.41   0.163    -.2341965    1.365867
         L1. |   .2145843   .5745405     0.37   0.710    -.9279526    1.357121
         L2. |   -.655617   .3925048    -1.67   0.099    -1.436156     .124922
             |
       _cons |   .0302675   .0666281     0.45   0.651    -.1022298    .1627648
------------------------------------------------------------------------------

. test (1 - L1.ln_inv - L2.ln_inv = 0) (ln_inc + L1.ln_inc + L2.ln_inc = 0)

 ( 1)  - L.ln_inv - L2.ln_inv = -1
 ( 2)  ln_inc + L.ln_inc + L2.ln_inc = 0

       F(  2,    84) =    3.08
            Prob > F =    0.0510

The p-value of the test command is of course invalid.